2005 AMC 10A Problem 17

Attempt Problem 17 of the 2005 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2005 AMC 10A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

17.

In the five-sided star shown, the letters A,A, B,B, C,C, D,D, and EE are replaced by the numbers 3,3, 5,5, 6,6, 7,7, and 9,9, although not necessarily in this order. The sums of the numbers at the ends of the line segments AB,AB, BC,BC, CD,CD, DE,DE, and EAEA form an arithmetic sequence, although not necessarily in this order. What is the middle term of the arithmetic sequence?
99
1010
1111
1212
1313
Answer: D
Concepts:arithmetic sequencedouble countingmean
Small hint:
Each of the five numbers is an endpoint of exactly two of the segments
Big hint:
The five segment sums total 2(3+5+6+7+9),2(3 + 5 + 6 + 7 + 9), and the middle term of a five-term arithmetic sequence is its mean
Solution:
Every number is an endpoint of two segments, so the five segment sums total 2(3+5+6+7+9)=60.2(3 + 5 + 6 + 7 + 9) = 60. The middle term of a five-term arithmetic sequence equals its mean, which is 605=12.\dfrac{60}{5} = 12.
Thus, the correct answer is D.
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